A student dilutes 11.5 mL of 6 M NaOH solution with enough water to make a total

Question

A student dilutes 11.5 mL of 6 M NaOH solution with enough water to make a total volume of 525 mL. What is the molarity of the diluted NaOH? (Pay attention to significant figures!) 25.00 mL of a standard solution of 0.1105 M HCI requires 27.31 mL of an NaOH solution to reach the phenolphthalein endpoint. What is the molarity of the NaOH solution? If an acid sample of mass 0.231 g required 38.22 mL of 0.09767 M NaOH to neutralize where 2.14 mL of 0.1015 M HCI was used to back titrate, find the equivalent mass of the unknown solid acid in grams per mole of H^+.

Explanation / Answer

1)

initial volume of NaOH = 11.5 mL

concentration = 6 M

final volume = 525 mL

M1 V1 = M2 V2

6 x 11.5 = M2 x 525

M2 = 0.131

Molarity of diluted NaOH = 0.13 M

2)

HCl   + NaOH   -------------> NaCl + H2O

M1 V1 = M2 V2

25 x 0.1105 = M2 x 27.31

M2 = 0.1012

Molarity of NaOH = 0.1012 M

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.